Population genetics and HW question

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MrMistery
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Population genetics and HW question

Post by MrMistery » Wed Jul 04, 2007 11:35 am

Another one which i haven't been able to solve... Again from IBO 2005

70% of the population of Beijing is able to taste phenylthiocarbamide. The ability to
taste (T, taster) is dominant over the inability to taste (t, non-taster). What percentage of the offspring of'tasters' will be non-tasters? (2 points)
A. 25%
B. 15%
C. 13%
D. 20%
E. 7.5%

First of all i don't get if it wants only the offspring of taster*taster or also taster*non-taster. For the sake of solving it, i have tried to solve it only for taster*taster. Here's what i did:
i first calculated the allele frequences with HW and i got p(T)=0.46 and q(t)=0.54. Then genotypes:
TT-20%
Tt-50%
tt-30%

And that is about all i am sure of. Then i started to do some math. i calculated from the total tasters how many TT and Tt there are, and got: 71% Tt and 28% TT. by crossing them i got a probability of 12,7% of getting a tt individual(non-taster).

Is it any good? Can anyone who has actually solved it run the whole thing past me so i actually understand it? Please...
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Re: Population genetics and HW question

Post by siroma » Wed Jul 04, 2007 2:30 pm

1. Assume HW equilibrium
f(t/t) = 0.3 => f(t) = sqrt(0.3)

2. f(t) = 0.5*f(T/t) + f(t/t) = 0.5*f(T/t) + 0.3 = sqrt(0.3)
f(T/t) = 2*(sqrt(0.3)-0.3)

3. f(T/T) = 0.7 - f(T/t)

Upto now:
0.20455488498967 T/T : 0.49544511501033 T/t : 0.3 t/t

4. Now I calculate frequency of t which comes from T/t and t which comes from t/t:

f(t(T/t)) = f(T/t)/2 = 0.24772255750517
f(t(t/t)) = f(t/t) = 0.3

5. We want frequencies of t/t, but only those who come from tasters:

(a) Tasters marry tasters
f(t/t) = f(t(T/t)) * f(t(T/t)) = 0.24772255750517^2 = 0.06136646549690

(b) Tasters marry non-tasters:
f(t/t) = 2*f(t(T/t))*f(t(t/t)) = 0.24772255750517 * 0.3 * 2 = 0.14863353450310

(c) Non-tasters marry non-tasters:
f(t/t) = f(t(t/t)^2 = 0.3^2
These gues we don't want, because they are not children of tasters

6. (a) + (b) = 0.06136646549690 + 0.14863353450310 = 0.21000000

Thus I choose answer (D)

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Post by siroma » Wed Jul 04, 2007 2:32 pm

By the way, can you upload somewhere questions from IBO ?

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Post by siroma » Wed Jul 04, 2007 2:46 pm

Btw, you can check my solution:
In HW equilibrium f(t/t) will be constant. So, if we add up (a) + (b) + (c) than it has to be 0.3.

Indeed, (a)+(b)+(c) = 0.21 + 0.3^2 = 0.21+0.09 = 0.3
and that's really so.

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Post by MrMistery » Wed Jul 04, 2007 5:32 pm

siroma, you just became my life-saver.
By the way, the questions from IBO 2003-2006 are hosted on this site, which is from the Tadjikistan olympiad.
http://bio.shelale.org/IBO.html
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Post by siroma » Wed Jul 04, 2007 6:11 pm

MrMistery wrote:siroma, you just became my life-saver.
By the way, the questions from IBO 2003-2006 are hosted on this site, which is from the Tadjikistan olympiad.
http://bio.shelale.org/IBO.html


Cool!
Thanks for link. ;)

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Post by MrMistery » Wed Jul 04, 2007 9:03 pm

ok, so i understant the your answer. However i have a question. Isn't it that there will be 21% out of the total number of offspring? Cause that is how i see it.
21% - percent of total offspring that are descendants of tasters and are non-tasters. But the question asks "What percentage of the offspring of'tasters' will be non-tasters?"...
Sorry, i am a little confused, and it is midnight here, so don't be too harsh if this doesn't make sense
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Post by siroma » Wed Jul 04, 2007 9:49 pm

I'm sorry. Your answer is right. I didn't take into account gametes of T/T tasters.

The answer is C.

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Post by siroma » Wed Jul 04, 2007 10:08 pm

Not only long solution, but so incorrect. :(

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Post by blcr11 » Thu Jul 05, 2007 3:49 am

At the risk of being totally and completely wrong, I offer an alternative. I’m not entirely convinced of my own reasoning, either, but I cleanly get one of the possible answers—no rounding off or anything.

There is only one type of mating between tasters that can yield non-tasters, namely, Tt x Tt. But the other types of crosses contribute to the total number of crosses between tasters, just not to the generation of tt individuals. The probability of generating tt individuals is the product of the probability of all taster matings times the probability of generating tt individuals from the particular crossings, or: Pr[matings of all tasters] x Pr[tt offspring from each type of cross].

Pr[of TTxTT mating] = 0.2x0.2 = 0.04
Pr[of TTxTt mating] = 0.2x0.5 = 0.10
Pr[of TtxTt mating] = 0.5x0.5 = 0.25
Pr[matings of all tasters] = 0.04 + 0.1 + 0.25 = 0.30

Pr[tt offspring from TTxTT cross] = 0
Pr[tt offspring from TTxTt cross] = 0
Pr[tt offspring from TtxTt cross] = 0.25
Pr[tt offspring from each type of cross] = 0 + 0 + 0.25

Pr[tt individual] = .3 x .25 = 0.075 or 7.5%.

Eh. I think I should stick to protein structure.

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Post by MrMistery » Thu Jul 05, 2007 7:43 am

Looks logical to me...
Siroma, how did you get 13%?

Now i know why i like biochemistry. None of this :D
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Post by siroma » Thu Jul 05, 2007 10:17 am

HW equilibrium:
q^2 = 0.3 =>
q = sqrt(0.3) = 0.55 ( freq of t )
p = 1 - 0.55 = 0.45 ( freq of T )

Now we'll get frequences of genotypes:
p^2 = 0.45^2 = 0.2 (T/T)
2*p*q = 2*0.45*0.55 = 0.5 (T/t)
q^2 = 0.55^2 = 0.3 (t/t)

Now we just cross all tasters:
T/T x T/T = 0.2 * 0.2 = 0.04 (all tasters)
T/T x T/t = 0.2 * 0.5 = 0.1 (all tasters)
T/t x T/T = 0.5 * 0.2 = 0.1 (all tasters)
T/t x T/t = 0.5 * 0.5 = 0.25 (1/4 non tasters)

All offspring = 0.04 + 0.1 + 0.1 + 0.25 = 0.49
Nontasters = 0.25 * 1/4 = 0.0625
0.0625 / 0.49 = 12.76 %

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